1. **Problem statement:** Show that $$\lim_{n \to \infty} \frac{1}{2n - 1} = 0$$ using the definition of the limit. 2. **Definition of limit for sequences:** For a sequence $(a_n)$, $$\lim_{n \to \infty} a_n = L$$ means that for every $$\epsilon > 0$$, there exists an integer $$N$$ such that for all $$n > N$$, $$|a_n - L| < \epsilon$$. 3. **Apply the definition:** Here, $$a_n = \frac{1}{2n - 1}$$ and $$L = 0$$. 4. We want to show that for every $$\epsilon > 0$$, there exists $$N$$ such that for all $$n > N$$, $$\left| \frac{1}{2n - 1} - 0 \right| = \frac{1}{2n - 1} < \epsilon.$$ 5. Solve inequality for $$n$$: $$\frac{1}{2n - 1} < \epsilon \implies 2n - 1 > \frac{1}{\epsilon} \implies n > \frac{1}{2\epsilon} + \frac{1}{2}.$$ 6. Choose $$N = \left\lceil \frac{1}{2\epsilon} + \frac{1}{2} \right\rceil$$. Then for all $$n > N$$, $$|a_n - 0| < \epsilon$$. 7. **Conclusion:** By the definition of limit, $$\lim_{n \to \infty} \frac{1}{2n - 1} = 0$$. This completes the proof for the first limit. **Note:** The user asked multiple limits but per instructions, only the first problem is solved completely.